Simplify; express your answer in exponential form. Assume $q\neq 0, x\neq 0$. $\dfrac{{(q^{-4})^{2}}}{{(q^{-1}x^{5})^{2}}}$
Explanation: To start, try working on the numerator and the denominator independently. In the numerator, we have ${q^{-4}}$ to the exponent ${2}$ . Now ${-4 \times 2 = -8}$ , so ${(q^{-4})^{2} = q^{-8}}$ In the denominator, we can use the distributive property of exponents. ${(q^{-1}x^{5})^{2} = (q^{-1})^{2}(x^{5})^{2}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(q^{-4})^{2}}}{{(q^{-1}x^{5})^{2}}} = \dfrac{{q^{-8}}}{{q^{-2}x^{10}}}$ Break up the equation by variable and simplify. $\dfrac{{q^{-8}}}{{q^{-2}x^{10}}} = \dfrac{{q^{-8}}}{{q^{-2}}} \cdot \dfrac{{1}}{{x^{10}}} = q^{{-8} - {(-2)}} \cdot x^{- {10}} = q^{-6}x^{-10}$.